1. ## inequality

when i have an inequality like this ,

$0<\sqrt{13}-3<1$

If i were to find $(\sqrt{13}-3)^4$ , can i say that

$0^4<(\sqrt{13}-3)^4<1^4 = 0<(\sqrt{13}-3)^4<1$ ??

And if the value at the side are negative , for instance

$
-5$

If i were find range of $x^2$ , am i right in saying .

$25>x^2>1$

Rearrange $1

2. Yes, you can

3. Originally Posted by songoku
Yes, you can

THanks .

How about only one of its side is negative ?

$-5

TO find $x^2$, do i do that this way :

$25>x^2>4$

Then rearrange $4

4. Originally Posted by thereddevils
THanks .

How about only one of its side is negative ?

$-5

TO find $x^2$, do i do that this way :

$25>x^2>4$

Then rearrange $4
x = 0 satisfies $-5. Does x = 0 satisfy $4 ....?

5. Originally Posted by mr fantastic
x = 0 satisfies $-5. Does x = 0 satisfy $4 ....?

THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .

6. Moral: do that as long as all pieces are positive.

7. Originally Posted by thereddevils
THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
The range of $f(x)= x^2$ is the set of all non-negative numbers.

8. Originally Posted by thereddevils
How about only one of its side is negative ?

$-5

TO find $x^2$, do i do that this way :

math]25>x^2>4 [/tex]

Then rearrange $4
Originally Posted by mr fantastic
x = 0 satisfies $-5. Does x = 0 satisfy $4 ....?
Originally Posted by thereddevils
THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
Draw a graph of y = x^2 for -5 < x < 2. Now look at the range of values of x^2.